The Normal distributionAQA A-Level Maths: Revision notes

The Normal distribution models a continuous quantity, such as a mass or a time, whose values cluster around a central value and tail away symmetrically. We write $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma^2$ is the variance, so $\sigma$ is the standard deviation. The graph of the probability density function is a symmetrical bell shape with its maximum at $x=\mu$, so the mean, median and mode are all equal to $\mu$. The total area under the curve is $1$, and the area between two values is the probability of lying between them. A histogram of continuous data (drawn with frequency density) from a Normal population is roughly bell-shaped. As the sample grows and the class widths shrink, the outline of the histogram approaches the curve. For a continuous variable $P(X=x)=0$, so $P(X<a)=P(X\le a)$.

Use the Normal cumulative distribution function on your calculator with the correct $\mu$ and $\sigma$ (not $\sigma^2$). Without the calculator, standardise: $Z=\frac{X-\mu}{\sigma}\sim N(0,1).$ Then use the standard Normal table or calculator. The curve is symmetrical, so $P(Z<-a)=1-P(Z<a)$ and $P(Z>a)=1-P(Z<a)$. Example: $X\sim N(60,4^2)$. Then $P(X>64)=P(Z>1)=1-0.8413=0.1587$ and $P(56<X<63)=\Phi(0.75)-\Phi(-1)=0.7734-0.1587=0.615$.

To find a value $x$ given a probability, use the inverse Normal function (or the table in reverse). If $P(X<x)=p$, then $x=\mu+z\sigma$ where $P(Z<z)=p$. For example, if $90\%$ of times are below $t$ when $T\sim N(25,6^2)$: $z=1.2816$, so $t=25+1.2816\times6=32.7$. When $\mu$ or $\sigma$ is unknown, standardise each given probability to get an equation. Two conditions give two simultaneous equations. For example, $P(Y<30)=0.2$ gives $30=\mu-0.8416\sigma$, and $P(Y>45)=0.1$ gives $45=\mu+1.2816\sigma$, so $\sigma=\frac{15}{2.1232}=7.07$ and $\mu=35.9$. A value below the mean has a negative $z$.

$\mu$ fixes the position of the curve, and $\sigma$ fixes its width. A larger $\sigma$ gives a wider, flatter curve, and a smaller $\sigma$ gives a narrower, taller one, because the total area stays $1$. The curve has points of inflection at $x=\mu-\sigma$ and $x=\mu+\sigma$, where the curve changes between curving upwards and curving downwards. For $N(60,4^2)$ these are at $56$ and $64$. Approximately $68\%$ of values lie within $1\sigma$ of the mean, $95\%$ within $2\sigma$ and $99.7\%$ within $3\sigma$. For example, in $N(500,8^2)$, $P(492<X<508)=0.683$. Comparing two distributions with the same mean, the one with the smaller $\sigma$ is more consistent.

A binomial variable $X\sim B(n,p)$ counts successes in $n$ independent trials. When $n$ is large and $p$ is not too close to $0$ or $1$, its bar chart is roughly bell-shaped and can be approximated by a Normal distribution with the same mean and variance: $X\sim B(n,p)\approx N\big(np,\;np(1-p)\big).$ A discrete variable is approximated by a continuous one, so use a continuity correction: $P(X\le45)\approx P(Y<45.5)$. Example: $X\sim B(100,0.4)$ has mean $40$ and variance $24$. Then $P(X\le45)\approx P\left(Z<\frac{45.5-40}{\sqrt{24}}\right)=P(Z<1.12)=0.869$. Conversely, a probability found from a Normal model can be used as the success probability in a binomial model, for example the number of saplings out of $5$ taller than $40$ cm, where each is taller with probability $p=0.283$.